Transformation of the Line-Element

[DrGreg]

Just to reinforce what DaleSpam is saying.

When you've chosen a coordinate system, the metric is a rank-2 tensor whose components (relative to that coordinate system) are$$g_{ab} = \left[\begin{array}{cccc} g_{00} & g_{01} & g_{02} & g_{03}\\ g_{10} & g_{11} & g_{12} & g_{13}\\ g_{20} & g_{21} & g_{22} & g_{23}\\ g_{30} & g_{31} & g_{32} & g_{33} \end{array}\right]$$When people write the equation$$ds^2 = g_{ab}\,dx^a\,dx^b$$that is just a convenient way of specifying what the components of the metric tensor are, without all the hassle of typesetting a 4×4 matrix. ds is the line element. g_ab is the metric tensor. They are not the same thing. One can be calculated from the other.

And its no use trying to define a 4D-coordinate system along a one-dimensional curve. For coordinate system to be valid (and to determine the components of the metric tensor relative to this coordinate system) it needs to be defined consistently over a 4-dimensional region of spacetime. (Or in general over an N-dimensional subset of an N-dimensional manifold.)

 

[Ben Niehoff]

Oh, certainly I think that Schwarzschild coordinates are special, because I've seen a common-sense argument^{http://www.mathpages.com/rr/s8-09/8-09.htm"} showing that

$$\left (\frac{\partial \tau}{\partial r} \right )^2\approx 1-\frac{G M}{r c^2}$$

The fact is, I have worked through this on my own time, and I know exactly what the variables are referring to. So when you make the claim that these variables are meaningless, you're telling me something I know is not true; at least for the g₀₀ component.

I disagree with your contention that you know what the variables are referring to. The weak-field limit calculation certainly does lead to the expression above for gravitational time dilation. But in the weak-field calculation, the variable 'r' is assumed to represent radial distance from the source.

In the full Schwarzschild metric, the coordinate 'r' no longer represents radial distance from the source, as clearly evidenced by the fact that $g_{rr} \neq 1$. However, it is still true that the gravitational time dilation, expressed in the coordinate 'r', takes the expression you have above.

If you were to express gravitational time dilation as a function of radial distance from the source, you would get a much more complicated expression! Only in the weak-field limit (i.e., very far from the source) would the expressions coincide.

Second, what I said originally (and I think I have continued to say) is that coordinates (in general) are not intrinsically meaningful. You have consistently ignored the qualifier "intrinsically".

I do agree that in specific coordinate systems we can often find meanings for the coordinates. But in order to do so, there is mathematical work to be done -- the meanings of the coordinates are not inherent or automatic. In fact, this work can be quite tough if we are given some unfamiliar metric, and it requires clear thinking rather than just making assumptions about what we think things mean. We have to look at geodesics, Killing vectors, integrate, measure, etc. Only then can we discover "This coordinate relates to this quantity".

For example, the Schwarzschild coordinate 'r' can be related to the circumferences of circles centered on the source point:

$$r = \frac{1}{2\pi} \oint_\mathcal{C} ds$$
But don't be fooled into thinking this implies that 'r' is a 'radius'! We are not in Euclidean space, so the usual circumference law of circles does not apply.

(P.S. What criteria of the Riemann curvature determines whether a manifold is flat?)

A manifold is flat if and only if the Riemann curvature tensor vanishes. Remember that if a tensor vanishes in one coordinate system, it vanishes in all coordinate systems.

Mathematically, the Schwarzschild metric identifies the transformation:

$$c^2 {d \tau}^{2} = \left(1 - \frac{2 G M}{r c^2} \right) c^2 dt^2 - \left(1-\frac{2 G M}{r c^2}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$

It's a differential equation, so you'll need some boundary values to define things explicitly.

But this isn't just a mapping of $$(dt,dr,d\theta,d\phi) \to ds$$

It is a mapping of $$(dt,dr,d\theta,d\phi) \to (d\tau,dR, d\theta',d\phi')$$

This is complete nonsense. A metric tensor is not a coordinate transformation. You seem to be extremely confused.

http://en.wikipedia.org/wiki/Coordinate_system In geometry, a coordinate system is a system which uses one or more numbers, or coordinates, to uniquely determine the position of a point or other geometric element.

The key word here is uniquely. We'll come back to that...

These spherical coordinates can be mapped to cartesian coordinates in the standard way

$$\begin{matrix} t=t\\ z=r \cos(\theta)\\ x=r \sin(\theta)\cos(\phi)\\ y=r \sin(\theta)\sin(\phi) \end{matrix}$$

Which is also a unique mapping, except at r=0, there are several different values of phi all mapping to the same point.

It would be instructive if you compute, and then post for us, the Schwarzschild metric after applying the above coordinate transformation. What does it look like? How should the (x,y,z) coordinate be interpreted? Is this the interpretation you thought they would have?

$$\begin{matrix} dR \equiv ds|_{d\theta=d\phi=dt=0}=\left ( \frac{1}{\sqrt{1-\frac{2G M}{r c^2}}} \right )dr\\ d\tau \equiv ds|_{d\theta=d\phi=dr=0}=\left ( \sqrt{1-\frac{2G M}{r c^2}} \right )dt \end{matrix}$$

(Note: I've corrected dr to dt in the second line, which is clearly what you meant. I've also put in some missing 2's.)

Look carefully here. As I tried to explain to Anamitra a long time ago, the second line is a lie. You have called a quantity $d\tau$ when it is not, in fact, an exact differential! This is not allowed.

In particular, the integral

$$\tau \; \; \overset{?}{=} \; \int_{t_0} \sqrt{1-\frac{2G M}{r c^2}} \; dt$$
does not give a unique answer, because the value of the integral depends on the path. This is because the quantity being integrated is not an exact differential. This is why I said earlier that $\tau$ is not a good coordinate.

Using boundary conditions of $$\begin{matrix} R(r=\frac{2G M}{ c^2})=0\\ \tau(t=0)=0 \end{matrix}$$

We can calculate the definite integrals:

$$\begin{matrix} R(r)=\int_{2G M/c^2}^{r}\frac{1}{\sqrt{1-\frac{2G M}{\rho c^2}}}d\rho\\ \tau(t,r)=\int_{0}^{t}\sqrt{1-\frac{2G M}{r c^2}}dt = \left (\sqrt{1-\frac{2G M}{r c^2}} \right )t \end{matrix}$$

Ah, now this is different. Here you have specified a path along which to do the previous integral. You have chosen the path $r = \mathrm{const}$. That's fine, I just want you to realize that you had to make that choice, and you could have made it another way.

Again, it would be instructive for you to compute, and post here, the Schwarzschild metric in these new coordinates. I gave the solution to your first integral in my previous post, if you want to use it:

$$\int_{2m}^r \Big( 1 - \frac{2m}{r'} \Big)^{-1/2} dr' = \sqrt{r ( r - 2m)} + 2m \; \cosh^{-1} \sqrt{\frac{r}{2m}}$$

If I am setting dr=0, I am saying, essentially there is no "drop" between the events. In a black hole, this might not be available; you'd have to be in a rocketship thrusting away from the center in order to maintain dr=0. But if you are on a solid planet, there is no problem. You just set your clock on a floor or table.

The correct term for this is a stationary observer. Note that not every spacetime has stationary observers! It turns out that the existence of a stationary observer is intimately connected to the existence of a timelike Killing vector (this is a fancy way of saying the spacetime has time translation symmetry). When there is a timelike Killing vector, we can call an observer stationary if his worldline is everywhere tangent to the timelike Killing vector.

 

[JDoolin]

Just to reinforce what DaleSpam is saying.

When you've chosen a coordinate system, the metric is a rank-2 tensor whose components (relative to that coordinate system) are $$g_{ab} = \left[\begin{array}{cccc} g_{00} & g_{01} & g_{02} & g_{03}\\ g_{10} & g_{11} & g_{12} & g_{13}\\ g_{20} & g_{21} & g_{22} & g_{23}\\ g_{30} & g_{31} & g_{32} & g_{33} \end{array}\right]$$ When people write the equation $$ds^2 = g_{ab}\,dx^a\,dx^b$$ that is just a convenient way of specifying what the components of the metric tensor are, without all the hassle of typesetting a 4×4 matrix. ds is the line element. g_ab is the metric tensor. They are not the same thing. One can be calculated from the other.

And its no use trying to define a 4D-coordinate system along a one-dimensional curve. For coordinate system to be valid (and to determine the components of the metric tensor relative to this coordinate system) it needs to be defined consistently over a 4-dimensional region of spacetime. (Or in general over an N-dimensional subset of an N-dimensional manifold.)

I'm still confused. Can you explicitly state what it is that I got confused, or what I got wrong, or what I'm "lying" about?

My impression is that
$$g_{ab} = \left[\begin{array}{cccc} g_{00} & g_{01} & g_{02} & g_{03}\\ g_{10} & g_{11} & g_{12} & g_{13}\\ g_{20} & g_{21} & g_{22} & g_{23}\\ g_{30} & g_{31} & g_{32} & g_{33} \end{array}\right]$$

is a rank 2 tensor, while $ds^2 = g_{ab}\,dx^a\,dx^b$ defines a scalar from a rank 2 tensor.

Is this wrong?

 

[Dale]

And its no use trying to define a 4D-coordinate system along a one-dimensional curve. For coordinate system to be valid (and to determine the components of the metric tensor relative to this coordinate system) it needs to be defined consistently over a 4-dimensional region of spacetime. (Or in general over an N-dimensional subset of an N-dimensional manifold.)

Yes, otherwise it is a mapping from the manifold to R1 instead of a mapping from the manifold to R4.

 

[Dale]

is a rank 2 tensor, while $ds^2 = g_{ab}\,dx^a\,dx^b$ defines a scalar from a rank 2 tensor.

Yes, this is correct. So how can you possibly conclude that therefore g is path-dependent?

 

[DrGreg]

I'm still confused. Can you explicitly state what it is that I got confused, or what I got wrong, or what I'm "lying" about?

My impression is that
$$g_{ab} = \left[\begin{array}{cccc} g_{00} & g_{01} & g_{02} & g_{03}\\ g_{10} & g_{11} & g_{12} & g_{13}\\ g_{20} & g_{21} & g_{22} & g_{23}\\ g_{30} & g_{31} & g_{32} & g_{33} \end{array}\right]$$

is a rank 2 tensor, while $ds^2 = g_{ab}\,dx^a\,dx^b$ defines a scalar from a rank 2 tensor.

Is this wrong?

I don't think there's anything wrong with your specific words above. The problem when you try to define a new coordinate by integrating a differential along a 1-D curve. That gives you a 1-D parameterisation of a 1-D curve, but in general there's no guarantee that this will extend to give you a self-consistent 4-D coord system over a 4-D region. I'll draw your attention to what I said earlier in another thread:

$${dq}{=}{R}{Sin}{(}{\theta}{)}{d}{\phi}$$

Because we must have$$dq = \frac{\partial q}{\partial \theta} d\theta + \frac{\partial q}{\partial \phi} d\phi$$your equation implies not only$$\frac{\partial q}{\partial \phi} = R \, \sin \, \theta$$but also$$\frac{\partial q}{\partial \theta} = 0$$As others have pointed out in different ways, there is no simultaneous solution to both those equations.

 

[Anamitra]

The Two-Sphere Problem:

We consider a Schwarzschild sphere with r=k [k: Consant]

Physically the radius of the sphere is given by:
$${R}{=}\int{(}{1}{-}\frac{2m}{r}{)}^{-1/2}{dr}{+}{C}$$ ------------ (1)
The integral extends forn r>2m to r=k, C is a constant
for any theta,phi direction. [Of course t=constant here ]Now in flat spacetime we take a sphere of radius=r=k

We use the transformation:
(R,theta,phi)------->(r,theta,phi) ------------- (2)
[The transformation is from Schwarzschild's space to flat space for t= some constant]
[The R on the LHS of (2) may be called the Physical radius of the Schwarzschild Sphere corresponding to the coordinate value of r=k]
The value of ds^2 is not changing on the surface of the two spheres.
Now what happens if we consider different spheres in each system for different values of k and use the same transformation[given by (2)] for each time slice?

 

[Dale]

The Two-Sphere Problem:

We consider a Schwarzschild sphere with r=k [k: Consant]

Physically the radius of the sphere is given by:
$${R}{=}\int{(}{1}{-}\frac{2m}{r}{)}{dr}$$ ------------ (1)
for any theta,phi direction. [Of course t=constant here ]

OK, so now we are doing a different transformation? So evaluating the integral we get:
$$R=r-2 m \log (r)$$
or solving for r we get
$$r=-2 m W\left(-\frac{e^{-\frac{R}{2 m}}}{2 m}\right)$$
where W is the product log function.

OK, so far so good. Are you hoping that this coordinate transform will allow you to flatten the metric?

 

[Anamitra]

I have done an editing on post 147

 

[Dale]

OK, so now we are back to post 126? If so, please see my previous response.

 

[Anamitra]

It is important to observe that some parallel issues are going on in this thread----and all of them are interesting.

1. The issue of projecting one curved surface [manifold] on another,particularly on a flat surface. Here the entire surface is in consideration.The value of ds^2[line element squared] changes.
ex: posting #1 ,Posting #2...

An interesting type of such a projection would be the consideration of two manifolds on the same coordinate grid.[The other issues are very much alive]
[#73,# 83,#111]


2. Projecting a specific path from one manifold to another. Here we may not consider the projection of the entire surface. Projection in the vicinity/neighborhood of the curve is sufficient. "ds^2" that is the line element squared changes.
ex:postings #126,#132

Issues 1 and 2 have a strong correlation---they are allied issues.If two manifolds stand on the same coordinate grid--we are in effect projecting one manifold on the other. Any curve in one manifold will have its projection on the other.



[In posting #111 some square root signs were left out inadvertently. This was mentioned at the end of #126]

[It is also important to note that I have used ds^2 to mean the square of the line element.i would continue with this meaning.If I use it in a different way it will be specified separately.]

The next issue,that is the third one belongs to an entirely different category

3. Passing from one manifold to another keeping "ds^2" invariant. "Flattening of a sphere" belongs in this category.Rather one may define it in this way.

I have been currently trying my hands with this [the third issue]with the posting #147, so far as the current thread is concerned.[The other issues are very much alive]

 

[Anamitra]

The flattening of a sphere could become somewhat easier in view of the fact that each one of the coefficients of the metric tensor may not be functions of all the four variables involved.This can be an advantageous feature.We may take for example Schwarzschild's Geometry.

 

[JDoolin]

Again, it would be instructive for you to compute, and post here, the Schwarzschild metric in these new coordinates. I gave the solution to your first integral in my previous post, if you want to use it:

$$\int_{2m}^r \Big( 1 - \frac{2m}{r'} \Big)^{-1/2} dr' = \sqrt{r ( r - 2m)} + 2m \; \cosh^{-1} \sqrt{\frac{r}{2m}}$$

I've done this integral numerically, and your solution appears to be incorrect. For instance, when I set m=1/2, it gives a negative value at r=1.1.

Here is what I calculated at some specific radii. 1.1, 1.01, 1.001, 1.0001, 1.0001

If I could compute a closed-form solution for the integral, that would be excellent, but I don't think you have it.

Some Typos have been corrected from the original posting

$$\int_{1}^{1.1}\frac{1}{\sqrt{1 -\frac{1}{r}}} = .6428 \Rightarrow \frac{\Delta R}{\Delta r} = 6.428$$

$$\int_{1}^{1.01} \frac{1}{\sqrt{1 -\frac{1}{r}}} = .2003 \Rightarrow \frac{\Delta R}{\Delta r} = 20.03$$

$$\int_{1}^{1.001} \frac{1}{\sqrt{1 -\frac{1}{r}}} = .0632 \Rightarrow \frac{\Delta R}{\Delta r} = 63.2$$

$$\int_{1}^{1.0001} \frac{1}{\sqrt{1 -\frac{1}{r}}} = .0200 \Rightarrow \frac{\Delta R}{\Delta r} = 200$$

$$\int_{1}^{1.00001} \frac{1}{\sqrt{1 -\frac{1}{r}}} = .00632 \Rightarrow \frac{\Delta R}{\Delta r} = 632$$

 

[Anamitra]

Unfortunately you get stuck even there. For the transforms you posted you do not get these expressions back. For example differentiating the R to r coordinate transform equation gives

$${R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}$$
$$dR=\frac{2 m \left(\frac{dr}{2 \sqrt{r-2 m}}+\frac{dr}{2 \sqrt{r}}\right)}{\sqrt{r-2 m}+\sqrt{r}}-\frac{2 r \, dr -2 m\, dr}{8 \left(r^2-2 m r\right)^{3/2}}$$
$$dR=\frac{m (4 r (r-2 m)+1)-r}{4 (r (r-2 m))^{3/2}}dr \ne \sqrt{{g}_{rr}}{dr}$$

That is the problem with trying to solve equations that don't have a solution. If you try and get some result then when you plug it back into the original equations you don't get what you thought you would. Please go ahead and do the differentiation yourself to confirm that I have done it correctly.

I am getting the value of dR as follows:

$${R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}$$

$${dR}{=}\frac{1}{2}\frac{2r-2m}{\sqrt{{r}^{2}{-}{2m}}}{dr}{+}{2m}\frac{1}{\sqrt{r}{+}\sqrt{r-2m}}{[}\frac{1}{{2}\sqrt{r-2m}}{+}\frac{1}{{2}\sqrt{r}}{]}{dr}$$
$${=}{[}\frac{r-m}{\sqrt{{r}^{2}{-}{2mr}}}{+}\frac{m}{\sqrt{{r}{(}{r}{-}{2mr}{)}}}{]}{dr}$$
$${=}\frac{r}{\sqrt{{(}{r}^{2}{-}{2mr}{)}}}{dr}$$
$${=}{(}{1}{-}\frac{2m}{r}{)}^{-1/2}{dr}$$
$${=}\sqrt{{g}_{rr}}{dr}$$

 

[Anamitra]

OK, so now we are back to post 126? If so, please see my previous response.

For a Schwarzschild sphere of radius r=k and an ordinary sphere radius=r=k the line element on each sphere is the same in length.[t=constant for both the spheres;ie time slices are being considered]

In each case the line element is given by:

$${ds}^{2}{=}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}{)}$$

In each case dr=0 and dt =0

The physical value of the radius of the Schwarzschild sphere is given by R in post #147

 

[Anamitra]

Thus in the last posting[post #155] we have flattened a sphere[r=constant ,t=constant].
One to one mapping from a Schwarzschild Sphere[ a sphere in curved space] to an ordinary sphere[a sphere in flat spacetime] exists for which the line element does not change in value.

 

[JDoolin]

The integration s requested in posting # 124


Let us consider a path:
Theta=const ,phi =const; r=kt where k is a constant
[An arbitrary path may not correspond to a geodesic]

$${dR}=\frac{dr}{\sqrt{{1}{-}\frac{2m}{r}}}$$
$$\int{dR}{=}\int\frac{dr}{\sqrt{{1}{-}\frac{2m}{r}}}$$
$$\int\frac{r}{\sqrt{{r}{(}{r}{-}{2m}{)}}}{dr}$$
$${=}\int\frac{1}{2}\frac{2r-2m}{\sqrt{{r}^{2}{-}{2mr}}}{dr}{+}{m }\int\frac{1}{\sqrt{{r}^{2}{-}{2mr}}}{dr}$$
$${R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}$$

Ah. Interesting. You split the integral into two solvable integrals.

The one on the left, you are doing substitution of

$$u=r^2-2 m r , du = 2 r - 2 m$$

So you simplify to
$$\int \frac{1}{2} u^{-1/2} du = \frac{1}{1/2}\frac{1}{2}u^{1/2} = (r^2- 2 m r)^{1/2}$$

I am not seeing how you got the other integral, though. [strike]And the final answer doesn't seem to square with the values I got by numerical integration.[/strike] (Edit: My mistake, both the numerical integration values, and this equation are correct.)

Again,
$${dT}{=}\sqrt{(}{1}{-}\frac{2m}{r}{)}{dt}$$
$${=}\sqrt{\frac{{r}{-}{2m}}{r}}{dt}$$
$${=}\sqrt{\frac{{kt}{-}{2m}}{kt}}{dt}$$
$${=}\frac{{kt}{-}{2m}}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{dt}$$
$${=}\frac{1}{2k}\frac{{2k}^{2}{t}{-}{2km}}{\sqrt{{k}^{2}{t}^{2}{-}{2ktm}}}{dt}{-}{m}\frac{1}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{dt}$$
$${Integral}{=}{T}{=}\frac{1}{k}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{-}\frac{2m}{k}{ln}{[}\sqrt{t}{+}\sqrt{(}{t}{-}{2m}{/}{k}{)}{]}{+}{C’}$$

The constants will be removed from initial conditions or by calculating definite integrals.

[In Post #111 the square root sign was inadvertently missed out in the expressions for dR and dT . I would request the audience to consider it.]

The only thing I would mention here is that by assigning a constant radial velocity, r=k t, you are trying to create a "coordinate system" full of objects that are moving in different directions, (radially inward or outward) which I'm pretty sure doesn't make sense. (or at least introduces a lot of unnecessary confusion)

When you set r= constant, that simplifies your calculation a lot, and it then represents a coordinate system defined by the positions of stationary clocks; both in terms of ∂R/∂τ = 0 and ∂r/∂t=0, if I'm not mistaken.

 

[Anamitra]

Ah. Interesting. You split the integral into two solvable integrals.

The one on the left, you are doing substitution of

$$u=r^2-2 m r , du = 2 r - 2 m$$

So you simplify to
$$\int \frac{1}{2} u^{-1/2} du = \frac{1}{1/2}\frac{1}{2}u^{1/2} = (r^2- 2 m r)^{1/2}$$

I am not seeing how you got the other integral, though. And the final answer doesn't seem to square with the values I got by numerical integration.

Its quite simple. We can do it in this way:
$$\int\frac {dr}{\sqrt{{r}^{2}{-}{2mr}}}$$
$${=}\int\frac {dr}{\sqrt{{r}{(}{r}{-}{2m}{)}}}$$

Let

$${r}{=}{z}^{2}$$
$${dr}{=}{2z}{dz}$$

Therefore,
$${Integral}{=}\int\frac{{2z}{dz}}{\sqrt{{z}^{2}{(}{z}^{2}{-}{2m}{)}}}$$
$${=}\int\frac{2dz}{\sqrt{{(}{z}^{2}{-}{2m}{)}}}$$
$${=}{2}{\,}{\,}{ln}{(}{z}{+}\sqrt{{z}^{2}{-}{2m}}{)}$$
$${=}{2}{\,}{ln}{(}\sqrt{r}{+}\sqrt{{r}{-}{2m}}{)}$$

 

[JDoolin]

$$\tau \; \; \overset{?}{=} \; \int_{t_0} \sqrt{1-\frac{2G M}{r c^2}} \; dt$$
does not give a unique answer, because the value of the integral depends on the path. This is because the quantity being integrated is not an exact differential. This is why I said earlier that $\tau$ is not a good coordinate.

Ah, now this is different. Here you have specified a path along which to do the previous integral. You have chosen the path $r = \mathrm{const}$. That's fine, I just want you to realize that you had to make that choice, and you could have made it another way.

Again, it would be instructive for you to compute, and post here, the Schwarzschild metric in these new coordinates. I gave the solution to your first integral in my previous post, if you want to use it:

$$\int_{2m}^r \Big( 1 - \frac{2m}{r'} \Big)^{-1/2} dr' = \sqrt{r ( r - 2m)} + 2m \; \cosh^{-1} \sqrt{\frac{r}{2m}}$$

The correct term for this is a stationary observer. Note that not every spacetime has stationary observers! It turns out that the existence of a stationary observer is intimately connected to the existence of a timelike Killing vector (this is a fancy way of saying the spacetime has time translation symmetry). When there is a timelike Killing vector, we can call an observer stationary if his worldline is everywhere tangent to the timelike Killing vector.

I'm not sure of all of the implications of this, but it seems to me, the way I've defined things, ∂τ/∂t is a function of r, while ∂τ/∂r= is a function of r and t.

More explicitly, I may have to set up a fairly complicated synchronization convention to get all the clocks set to internal coordinate time τ=0 at the external coordinate time t=0. But before that instant, and after that instant, the clocks aren't synchronized. (unless they are at the same radius) The clocks further from the center will be going faster than the clocks nearer the center.

So as time passes, the value of ∂τ/∂r at any given region will shrink and shrink before the synchronization convention, and will grow an grow afterward. This is no big deal. Any event is still uniquely located by (R,τ,θ,Φ) just as it is uniquely located by (r,t,θ,Φ). Each clock only passes through each time only once. No clocks are caused to go backward in time, and repeat an event over again.

I'm not an expert on Killing vectors, but I'm not sure you'd have translation symmetry in time. If you move 10 seconds into the future or the past, the clocks below you would lose time, and the clocks above you would gain time. So you could easily calculate how long until, or how long since the synchronization convention, based on looking at nearby clocks. But the relative rate between them would still be the same--so there wouldn't be any difference in the physics.

 

[Dale]

I am getting the value of dR as follows: ...
$${=}\sqrt{{g}_{rr}}{dr}$$

You are correct, I must have made a typo entering the equation in yesterday .

I will continue working with the rest of the transformation later today. It will not change the end result that the coordinate transformation cannot flatten a curved metric.

 

[JDoolin]

Its quite simple. We can do it in this way:

Excellent. Thank you. Now I realized that my spreadsheet defaults to Log base 10, rather than natural log. And I have verified that your equatin gives the same results as the numerical integration.

 

[PAllen]

Thus in the last posting[post #155] we have flattened a sphere[r=constant ,t=constant].
One to one mapping from a Schwarzschild Sphere[ a sphere in curved space] to an ordinary sphere[a sphere in flat spacetime] exists for which the line element does not change in value.

This may actually be true, but not very significant. Two surfaces of many geometries and topologies are embeddable in 3-space with the induced metric on them capturing geometry of a highly non-euclidean 2-manifold. I suspect that almost all two manifolds are embeddable in flat euclidean 4-space (certainly many that can't be embedded in flat 3-space can be embedded in flat 4-space). Thus the idea that pure 2-sphere (as a geometric as well topologic manifold) can be embedded in both Scwharzschild manifold and flat euclidean 4-space is a triviality with no significance the nature of mappings between the two 4-manifolds.

You haven't flattened anything. You've just established a well known feature of embedding: a flat space can embed lower dimensional curved manifolds; a curved manifold of one geometry can embed lower dimensional manifolds of all different geometries.

 

[JDoolin]

You haven't flattened anything. You've just established a well known feature of embedding: a flat space can embed lower dimensional curved manifolds; a curved manifold of one geometry can embed lower dimensional manifolds of all different geometries.

I don't think that's Anamitra's intention, or at least not mine, anyway. The point is that a coordinate system defined by stationary clocks in the schwarzschild coordinates can be mapped unambiguously to a coordinate system of stationary clocks in a cartesian coordinate system; event for event, a 1 to 1 relationship.

This is an embedding, or, if you prefer, a projection from one 4D space-time to another 4D spacetime, which could easily be co-located; for instance, simply by having the stationary clocks in the region each keep track of two different times.

 

[Anamitra]

Another Example of Flattening
a Curved Space-time Object
[-----The r-Slices "Flattened"]

We start with the Schwarzschild Metric:

$${ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}{)}$$ --------------- (1)
For Constant r we have:

$${ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}{)}$$ --------------- (2)

For the surface r=Constant,we use the substitution:

$${dT}{=}{(}{1}{-}\frac{2m}{r}{)}^{1/2}{dt}$$

$${T}{=}\int{(}{1}{-}\frac{2m}{r}{)}^{1/2}{dt}$$
the limits of integration extending from to t0 to t. This eliminates the constant of integration.
$${T}{=}{(}{1}{-}\frac{2m}{r}{)}{(}{t-t0}{)}$$For r=Const we have,the metric has the form:
$${ds}^{2}{=}{dT}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{Sin}^{2}{(}{\theta}{)}{d}{\phi}^{2}{)}$$ ----------- (3)

This is again a flat Spacetime metric[the above one--relation (3)]. "ds" remains invariant in the transformation used.
This time we have flattened a three dimensional surface[holding r as constant leaves us three independent variables--t,theta and phi]

[We have not changed theta and phi in our transformation]

 

[Dale]

a coordinate system defined by stationary clocks in the schwarzschild coordinates can be mapped unambiguously to a coordinate system of stationary clocks in a cartesian coordinate system; event for event, a 1 to 1 relationship.

Sure. That is what any coordinate chart does. They all perform an unambiguous 1 to 1 mapping from some open subset of the manifold to some open subset of R(N). That is part of the definition of a coordinate system, and is not in doubt.

The question is whether you can flatten the metric by a coordinate transform, which you cannot. And which despite lots of posts Anamitra has never demonstrated. In fact, Anamitra has never even bothered to actually take one of his transforms and calculate the curvature tensor, always simply assuming flatness.

 

[Dale]

Hi Anamitra,

For:
$$T=c+\frac{\sqrt{k t (k t-2 m)}}{k}-\frac{2 m \log \left(\sqrt{t-\frac{2 m}{k}}+\sqrt{t}\right)}{k}$$
and
$$r=k t$$
I am getting
$$\text{dT}=\frac{\frac{\text{dr} \left(4 m^2 t+m \left(2 r \sqrt{t} \sqrt{t-\frac{2 m t}{r}}-2 t \sqrt{r (r-2 m)}-2 r t\right)\right)+2 \text{dt} r (2 m-r) \left(\sqrt{r (r-2 m)}-m\right)}{2 m-r}+4 m (\text{dr} t-\text{dt} r) \log \left(\sqrt{t-\frac{2 m t}{r}}+\sqrt{t}\right)}{2 r^2}$$

 

[Anamitra]

If the screen could be normalized to the correct width...This is a request.

Hi Anamitra,

For:
$$T=c+\frac{\sqrt{k t (k t-2 m)}}{k}-\frac{2 m \log \left(\sqrt{t-\frac{2 m}{k}}+\sqrt{t}\right)}{k}$$
and
$$r=k t$$
I am getting ...

Lets do it:

$${Integral}{=}{T}{=}\frac{1}{k}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{-}\frac{2m}{k}{ln}{[}\sqrt{t}{+}\sqrt{(}{t}{-}{2m}{/}{k}{)}{]}{+}{C’}$$

Or,

$${dT}{=}\frac{1}{k}\frac{1}{2}\frac{{2}{k}^{2}{t}{-}{2mk}}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}{dt}{-}\frac{2m}{k}\frac{\frac{1}{{2}\sqrt{t}}{+}\frac{1}{\sqrt{{t}{-}{2m}{/}{k}}}}{\sqrt{t}{+}\sqrt{{t}{-}{2m}{/}{k}}}{dt}$$

$${dT}{=}\frac{kt-m}{\sqrt{{kt}{(}{kt}{-}{2m}}}{-}\frac{m}{k}\frac{1}{\sqrt{{t}{(}{t}{-}\frac{2m}{k}{)}}}$$

$${dT}{=}\frac{kt-m}{\sqrt{{kt}{(}{kt}{-}{2m}}}{-}\frac{m}{\sqrt{{kt}{(}{kt}{-}{2m}{)}}}$$

We have,

$${dT}{=}\frac{kt-2m}{\sqrt{{kt}{-}{2m}}}$$

$${dT}{=}\sqrt{\frac{kt-2m}{kt}}$$

But r=kt

Therefore we have,finally,

$${dT}{=}\sqrt{\frac{r-2m}{r}}{dt}$$

Or,
$${dT}{=}{(}{1}{-}\frac{2m}{r}{)}^{1/2}{dt}$$

 

[Ben Niehoff]

$$\int_{2m}^r \Big( 1 - \frac{2m}{r'} \Big)^{-1/2} dr' = \sqrt{r ( r - 2m)} + 2m \; \cosh^{-1} \sqrt{\frac{r}{2m}}$$

I've done this integral numerically, and your solution appears to be incorrect. For instance, when I set m=1/2, it gives a negative value at r=1.1.

Something is wrong with your numerics. I double-checked my antiderivative in Mathematica. See the attached image of the output.

 

[Dale]

But r=kt

Therefore
$$dk = \frac{\text{dr} t-\text{dt} r}{t^2} \ne 0$$

 

[Ben Niehoff]

I'm not sure of all of the implications of this, but it seems to me, the way I've defined things, ∂τ/∂t is a function of r, while ∂τ/∂r= is a function of r and t.

More explicitly, I may have to set up a fairly complicated synchronization convention to get all the clocks set to internal coordinate time τ=0 at the external coordinate time t=0. But before that instant, and after that instant, the clocks aren't synchronized. (unless they are at the same radius) The clocks further from the center will be going faster than the clocks nearer the center.

So as time passes, the value of ∂τ/∂r at any given region will shrink and shrink before the synchronization convention, and will grow an grow afterward. This is no big deal. Any event is still uniquely located by (R,τ,θ,Φ) just as it is uniquely located by (r,t,θ,Φ). Each clock only passes through each time only once. No clocks are caused to go backward in time, and repeat an event over again.

I'm not sure if you read me very carefully. I agree with you that once you make a definite choice of path for the $\tau$ integral, you end up with a valid coordinate system. In fact, the path $r = \mathrm{const}$ is rather convenient, because then the $\tau$ coordinate you get represents the elapsed proper time of stationary observers. Look up LeMaitre coordinates; he does something similar where he chooses the proper time of stationary observers as his time coordinate.

My point was that you could have chosen some other path, such as $r = kt$, and it would give you an entirely different definition of the new coordinate $\tau$. So it would be ambiguous to say

$$\tau = \int \sqrt{1-\frac{2m}{r}} \; dt$$
because no path has been specified. And it would be an outright lie to say

$$d\tau = \sqrt{1-\frac{2m}{r}} \; dt$$
because the right-hand side is not a total differential.

I'm not an expert on Killing vectors, but I'm not sure you'd have translation symmetry in time. If you move 10 seconds into the future or the past, the clocks below you would lose time, and the clocks above you would gain time. So you could easily calculate how long until, or how long since the synchronization convention, based on looking at nearby clocks. But the relative rate between them would still be the same--so there wouldn't be any difference in the physics.

If you don't understand the significance of Killing vectors, haven't computed a curvature tensor, and don't know how to solve "fancy differential equations" to obtain the easiest GR solutions (Schwarzschild and Reissner-Nordstrom), then I would say you have some woodshedding to do.

Please continue to study Carroll; it's an excellent book, and it will surely clear up all these misunderstandings.

 

[Anamitra]

Therefore
$$dk = \frac{\text{dr} t-\text{dt} r}{t^2} \ne 0$$

K is a constant--that was mentioned in the relevant posting

The integration s requested in posting # 124Let us consider a path:
Theta=const ,phi =const; r=kt where k is a constant
[An arbitrary path may not correspond to a geodesic]

You could view it in this manner also[remembering k=constant]:

r=kt
dr=kdt
tdr-rdt=tkdt-ktdt=0

 

[Dale]

K is a constant--...

r=kt

Since this is a coordinate transform you can't have it both ways. Either k is a constant or r=kt. Pick one, they contradict each other. I honestly don't care which you pick.

 

[Anamitra]

Since this is a coordinate transform you can't have it both ways. Either k is a constant or r=kt. Pick one, they contradict each other. I honestly don't care which you pick.

In Post #126 we were looking for a coordinate transformation along the path stated below[in relation to Schwarzschild's Geometry]:
r=kt; k: constant
theta=constant
phi=constant

There is absolutely no contradiction in having k as constant in the path r=kt

 

[Dale]

There is absolutely no contradiction in having k as constant in the path r=kt

Yes, there is. See the bottom figure on page 36 and the top figure on page 37 of the Carroll notes with the accompanying text.

A coordinate system is a one-to-one invertible map from open subsets of the manifold to open subsets of R(n). The set defined by r=kt is not an open subset for a constant k. Therefore, if you allow both k constant and r=kt then you do not have a coordinate transformation. This was already mentioned by Dr. Greg in 141, and emphasized by me in 144.

So as I said before, since this is a coordinate transform you cannot have it both ways, either k is a constant or r=kt. You must pick one because they contradict each other.

 

[Anamitra]

Yes, there is. See the bottom figure on page 36 and the top figure on page 37 of the Carroll notes with the accompanying text.

A coordinate system is a one-to-one invertible map from open subsets of the manifold to open subsets of R(n). The set defined by r=kt is not an open subset for a constant k. Therefore, if you allow both k constant and r=kt then you do not have a coordinate transformation. This was already mentioned by Dr. Greg in 141, and emphasized by me in 144.

So as I said before, since this is a coordinate transform you cannot have it both ways, either k is a constant or r=kt. You must pick one because they contradict each other.

The basic issue here is to have open subsets on the line in question[and the corresponding map should comprise open subsets]:The line has to be represented as the union of open subsets.

r=kt; where k is constant
theta=const
phi=const

Now we may think of small portions of the above mentioned line which are open subsets.These portions [small segments into which we will divide the line:These segments may overlap]should not include their end points[the limit points]

We simply express the the line as the union of open subsets.

In fact any open subset should not contain its limit points. The line I have defined falls in this category.

If an open subset is 3D region then the surface containing its limit points is two-dimensional
If an open subset is 2D region then the set containing its limit points is one-dimensional,that is a curve.
If an open subset is 1D region then the set containing its limit should contain two distinct points.